Chapter 16, Problem 119A

Interpretation Introduction

Interpretation: The volume of oxygen that is required to completely burn $35\text{g}$ of methanol is to be calculated.

Concept introduction: For ideal gases at STP condition the volume occupied by one mole of it is 22.4L.

The expression for moles is as follows:

  $\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}$   ...... (1)

Answer to Problem 119A

The volume of oxygen required to completely burn $35\text{g}$ of methanol, ${\text{CH}}_{\text{4}}\text{O}$ is $36.5\text{L}$ .

Explanation of Solution

The given amount of methanol, 36.5L is $35\text{g}$ .

The molar mass of methanol is

  $\begin{array}{c}{\text{CH}}_{\text{4}}\text{O}=12+4+16\\ =32\end{array}$

Substitute $35\text{g}$ for mass of ${\text{CH}}_{\text{4}}\text{O}$ and $32\text{g/mol}$ for molar mass of ${\text{CH}}_{\text{4}}\text{O}$ in equation (1).

  $\begin{array}{c}\text{moles}\left({\text{CH}}_{\text{4}}\text{O}\right)=\frac{\text{ 35 g}}{\text{32 g/mol}}\\ =1.09\text{ mol}\end{array}$

From the equation,

  $2{\text{CH}}_{\text{4}}\text{O}\left(l\right)+3{\text{O}}_2\left(g\right)\to 2{\text{CO}}_{\text{2}}\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)$

For 2 moles of methanol, about 3 moles of oxygen is required.

Therefore, for one mole of methanol, $\frac{3}{2}$ moles of oxygen is required.

So, $1.09\text{ mol}$ of methanol needs

  $1.09\text{ mol}\times \frac{3}{2}=1.63molofoxygen$

The number of moles of oxygen required to burn $35\text{g}$ of methanol is $1.63\text{ mol}$ .

One mole of an ideal gas occupies a volume of about $22.4L$ .

So, $1.63\text{ mol}$ of oxygen needs,cbv

  $1.63\times 22.4=36.5L$

Conclusion

Therefore, the volume of oxygen required to completely burn $35\text{g}$ of methanol is $36.5\text{L}$ .5L.